This lecture is going to be about 2-dimensional projectile motion, and the first thing we're going
to do is define those two words. So for something to be 2-dimensional means that it's happening on
two perpendicular axes at the same time. Dimension is an axis of motion that you can move on and that
is at a 90 degree angle to other axes. So so far we've been talking about motion that happens in
just 1 dimension, that means back and forth along a straight line, but it's possible for
motion to happen in 2 dimensions at the same time, where it's moving up and down and left and right,
and it's also obviously possible for motion to happen in 3 dimensions, which is something
that we'll talk about in future units, but for now we want to focus on 2-dimensional motion
specifically. And a projectile is any object with only the force of gravity acting on it
in freefall. We talked about this a little bit in the acceleration unit when we talked about falling
objects. So as long as this object is in the air, it's a projectile, and as long as the only force
acting on it is gravity. it's a projectile. So 2-dimensional projectile motion means movement
of an object in both a horizontal and vertical direction where gravity is the only force present.
As it turns out this is a really important topic to understand in physics because a lot of objects
in the world fit this description. So if we watch this ball being launched, you can see on my y-axis
that's moving up and down in the y axis and it's moving from left to right on the x axis as well,
so it's 2-dimensional motion because it's moving along both axes. So the forces on
a projectile are a little counterintuitive. If we draw the chart of the forces at different points,
it's always only going to be the force of gravity. There is no other force acting on a projectile.
Later on in this unit we'll talk about force of air resistance that can act on projectiles,
but for most physics problems we assume that air resistance is not there, so the only other
force that can be acting on this projectile is the force of gravity. The reason why it's still
able to move from left to right is because of Newton's First Law. Newton's First Law says
that if there's no force on an axis, that means there's no change in velocity along that axis,
so if this projectile starts with some x-velocity it's gonna continue having that x-velocity through
the entire trip. So if it's moving to the right at 5 meters per second, it's gonna continue to
do that, because there's no force in either direction to stop it. The acceleration of a
projectile is always 9.81 straight down because again the only force acting on it is gravity,
and gravity on earth accelerates everything at 9.81 meters per second down, and the velocity
of a projectile is the way that it's actually moving, the direction of its motion. We can break
this into its x-velocity and y-velocity, how fast it's going in the x-direction and how fast it's
going in the y-direction, and the x-velocity is going to stay the same the whole time. If you're
able to find a projectile's x-velocity at any one point along its motion, you'll know that that's
the same velocity for all other points along the projectile's motion, because there's no force to
speed it up or slow it down in that direction. So that means that the x-velocity for a projectile is
going to stay the same through the whole motion. The y-velocity however is being changed, because
there's a downward acceleration of gravity, so for my example here this particular projectile
starts with a very positive upward velocity, but as it rises it slows down in the upward direction,
because acceleration is taking away from it. It's being added to in the negative direction at the
very top of its motion. You may remember from the unit on acceleration that there's no y-velocity at
the very top, because what's happening at the very top of its motion is that it's slowing to a stop,
and then beginning to fall again in the other direction, and then as it falls it starts to
get faster and faster in the negative direction because gravity is still accelerating it down.
So these are the x and y-velocities. And to understand what the overall velocity looks like
I just need to do vector addition by combining the tip of one vector to the tail of the other
and then drawing that new connecting vector, and these are the resulting velocities at each point,
which kind of fits our experience of a projectile. It starts going up at a large angle but then slows
until it's only moving perfectly horizontally and then begins to fall again. So this is what
the velocity of a projectile tends to look like. Okay. this sentence is the fundamental idea of 2D
projectile motion, this is kind of the guiding idea that we use to solve most of the problems
in this unit: Kinematic variables on one axis are independent of kinematic variables on the other
axis, except for time. We can use a separate set of kinematic variables for each axis to predict
the object's motion along that axis. So if I look at what the object is doing along the y-axis
versus what it's doing along the x-axis, there's no clear relationship there, like it's moving
up and then down along the y-axis but it's only moving rightward along the x-axis, so these two
things are independent of each other. So we can come up with 2 separate kinematic tables: one for
the x axis and one for the y axis, where we find missing information about the object. So we're
going to begin every single 2D projectile problem by making these tables for the x and y kinematic
variables, and again notice that time doesn't have a direction. It will be the only variable that is
the same in both. These are some general rules for how to solve a 2D projectile motion problem:
because gravity is the only force on the object, the x-direction acceleration will always be zero,
and the y-direction acceleration will always be 9.8 meters per second down. Whether you consider
down to be positive or negative is going to depend on what you want to do in that particular
problem. You just need to make sure that it is consistent with your other variables. The second
thing to know is that you can only use kinematic equations on one table at a time. At no point can
you put x-axis variables and y-axis variables together into a single kinematics equation,
so you can only work with one table at a time in these problems, and time is the only variable that
is the same in both, and if you don't have enough variables on one table you should solve for time
on the other table and bring it over to the table that's missing variables. You'll remember that
you need at least 3 kinematics variables to use the kinematics equations, so if you only have
2 on one table and you don't have time you can solve for time on the other table, bring it over
and solve some problems with it. I'll give you a few examples of that in just a moment. These are
some code words for projectile motion: "launched horizontally" or "falls off a flat surface" means
that the object has no initial y-velocity, or the y-velocity is zero. The highest point in its
motion or the maximum height it reaches is where the y-velocity is equal to 0, so that means that
if you're dealing with the highest point that an object reaches, that is the point where the final
velocity of the object is 0 meters per second in the y-direction, not in the x-direction. The
x-direction stays the same, but in the y-direction it's 0. The range is the total x displacement of a
projectile, and if the object lands at the same height it started with its y-displacement is
going to be 0, because if it goes up and then down it's at the same place where it started,
so its total displacement is 0. Let's use these concepts to solve a few problems: A ball rolls off
a 1.5 meter high table at 2 meters per second. How far away from the table will it land? So that's
the ball's motion and we're trying to figure out how far away from the table will it land. I'm
gonna call rightward my positive direction on my x-axis, and I'm going to call downward my positive
direction in the y-axis, just because this object isn't moving up at all, it's only falling down,
so it'll be easier to call down positive. And I know that because there's no force in the
x-direction the acceleration in the x-direction is 0, and in the y-direction is 9.8. This is true for
all 2D kinematics problems, and here I'm calling down positive, so that 9.8 in my y-direction is
positive, and the goal of the problem is to find how far away it is from the table which is its x
displacement, is how far it's moved horizontally. Vertically I can see that it's moving 1.5 meters,
but horizontally I don't know how far it moves, so that's the thing I'm trying to solve for. I
can also see that in the X direction it's moving at 2 meters per second perfectly horizontally,
so that means that it has a horizontal x-axis velocity of 2 meters per second, and because
that's pointing to the right I'll call that positive and put that in the initial velocity,
and in the y part it's moving down at 1.5 meters of displacement, so because I'm calling down
positive I'm calling that 1.5 meters positive as well, and because it's moving in the perfectly
horizontal direction to start off that means that it's not moving up or down at all in the
moment when it starts its projectile motion. So if it's not moving up or down at all to start off
that means that its initial y-velocity is going to be 0. That's all the information that we can get
from the beginning of this projectile problem. Most of the challenge of solving projectile
motion problems really comes from having to read out that information and understand what it's
telling you. After you do that the rest is pretty straightforward, you just look for what you're
trying to look for, and I can see that I'm trying to find the displacement in the x-direction,
and I can see that I only have 2 other variables in the x-direction, so I don't have enough in that
table to solve for the displacement. So what I can do is try to solve for time on the y-axis table
and then bring time over because the time that it falls is going to be equal to the time that it's
in the air in the x-direction as well. So looking at my y table I see I have s u a and I need t, and
the only thing that connects those is this second kinematics equation. So when I plug in my numbers
I get t by itself is equal to 0.56 seconds, so that's the total time that it's in the air before
it falls the full 1.5 meters. So that means that it's also the time that it's in the air in the
x-direction. So I can take that over into the x table and now I can solve for the displacement,
and I can see I have s u a and t so I'm actually going to use that same equation, but this time
I'm solving for s, and when I do that I get 1.12 meters, so that means that if you launch a ball
off a 1.5 meter table at 2 meters per second and there's no air resistance it will go exactly 1.12
meters away from the table. This is example number 2: a ball is launched at 10 meters per second at
a 30 degree angle above the horizontal. How far away will it land? So this is the motion of the
ball and we're trying to figure out how far away it lands. I'm gonna call the rightward
direction on the x-axis positive and the upward direction on the y-axis positive. You can choose
which directions are positive and negative on your own, that's okay as long as they're consistent.
Just like with all 2D projectile problems, I know that the acceleration in the x-direction is 0 and
the acceleration in the y-direction is 9.8 down, which because up is positive here I'm gonna have
to call down negative, so that acceleration is negative. That's gonna be really important for
solving these problems, and I can see that I have a problem where the initial velocity isn't in the
x-direction or the y-direction, it's in both it's moving both horizontally and vertically at the
same time, so to figure out how fast it's going in the x-direction and y-direction I can find the
x-component and y-component of that velocity using trig. One more thing I can see is that the total
displacement in the y-direction is zero meters because it's going up but then falling back down
to its original position, so it didn't actually move in total in the up down direction from the
place where it started, so its total displacement in the y-direction is 0. Solving for those initial
velocities in the x and y direction using trig gets me these answers: in the X direction it's
8.7 meters per second and in the y-direction it's 5 meters per second up which is positive,
and if I'm trying to figure out how far away it will land I need to find the displacement in the
x-direction. Just like last time we don't have enough variables in the x-direction to solve for
the displacement, but we can solve for time on the y-axis and then take it over to the x-axis and use
that to solve the problem. So I'm gonna start by solving for time, and I have the s u a and t, so
just like the previous problem we're gonna use the same equation, and when we do that we get time is
equal to 1.202 seconds, and because time is always the same in both axes the time in the y-direction
is equal to the time in the x-direction, so I can use that and the same equation to solve for
the displacement, and when I do that I get that the total displacement of the ball is 8.9 meters.
Here's one more example: a ball is launched at 10 meters per second at a 30-degree angle,
same as before, above the horizontal. What will be its velocity after 0.3 seconds? So I'm keeping
the same variables that I knew before, only this time it's not going to land, we're not analyzing
the spot where it lands yet, we're analyzing a previous spot. So if I'm trying to find its
velocity, that means its total velocity, not just its velocity in the x-direction or the
y-direction. So to solve for that I'm gonna have to find each individual x & y velocity and then
put them together to find the total, so I'll start in the x-direction. Looking in the x-direction I
have u a t and I need v, so that's going to be the first equation, and when I solve for that I
get the final is equal to 8.7 meters per second, which makes sense because if the acceleration is
0 the velocity isn't changing in the x-direction, and in the y-direction if I plug in my values for
u a and t into that same equation I get final v is equal to 2.1 meters per second. So if it moves
in the air 4.3 seconds, this is how fast it's going in the x-direction and the y-direction,
so now that I have the x-direction velocity and the y-direction velocity of the projectile when
it's been moving in the air 4.3 seconds. At that point right there I can figure out the total
velocity in this way. I know that the X velocity is this much and the Y velocity is this much and
these are two individual velocity vectors, the total velocity is going to be the vector sum
of the two which means a single new vector drawn from the tail of the first to the tip of the last
which looks like this, and when I plug those into the Pythagorean Theorem (because this is forming
a right triangle) I find that the hypotenuse of that triangle is 8.9 meters per second so that
would be the ball's total velocity after that amount of time. So those are a few examples
of solving projectile motion problems. You're going to get a lot more as the unit goes on.